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Quantum Theory and the Electronic Structure of Atoms

Quantum Theory and the Electronic Structure of Atoms. Chapter 6 . Properties of Waves. Wavelength ( l ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Properties of Waves.

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Quantum Theory and the Electronic Structure of Atoms

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  1. Quantum Theory and the Electronic Structure of Atoms Chapter 6

  2. Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough.

  3. Properties of Waves Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n

  4. Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c

  5. A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? l n Radio wave l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m l = 5.0 x 1012 nm

  6. Mystery #1, “Black Body Problem”Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x 10-34 J•s

  7. Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905 hn • Light has both: • wave nature • particle nature KE e- Photon is a “particle” of light hn = KE + BE KE = hn - BE

  8. When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. E = h x n E = h x c / l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J

  9. Line Emission Spectrum of Hydrogen Atoms

  10. ( ) En = -RH 1 n2 Bohr’s Model of the Atom (1913) • e- can only have specific (quantized) energy values • light is emitted as e- moves from one energy level to a lower energy level n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J

  11. E = hn E = hn

  12. ni = 3 ni = 3 f i ni = 2 nf = 2 DE = RH i f ( ) ( ) ( ) Ef = -RH Ei = -RH nf = 1 nf = 1 1 1 1 1 n2 n2 n2 n2 Ephoton = DE = Ef - Ei

  13. Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. Ephoton = DE = RH i f ( ) 1 1 n2 n2 Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l=1280 nm

  14. Why is e- energy quantized? De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nll = h/mu u = velocity of e- m = mass of e-

  15. What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? l = h/mu h in J•s m in kg u in (m/s) l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6) l = 1.7 x 10-32 m = 1.7 x 10-23 nm

  16. Schrodinger Wave Equation • In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- • Wave function (Y) describes: • . energy of e- with a given Y • . probability of finding e- in a volume of space • Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

  17. n=1 n=2 n=3 Schrodinger Wave Equation Y = fn(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus

  18. Where 90% of the e- density is found for the 1s orbital e- density (1s orbital) falls off rapidly as distance from nucleus increases

  19. Schrodinger Wave Equation Y = fn(n, l, ml, ms) angular momentum quantum number l for a given value of n, l= 0, 1, 2, 3, … n-1 l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e- occupies

  20. l = 0 (s orbitals) l = 1 (p orbitals)

  21. l = 2 (d orbitals)

  22. Schrodinger Wave Equation Y = fn(n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or2 orientation of the orbital in space

  23. ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2

  24. Schrodinger Wave Equation Y = fn(n, l, ml, ms) spin quantum number ms ms = +½or -½ ms = +½ ms = -½

  25. Schrodinger Wave Equation Y = fn(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Y. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time

  26. How many electrons can an orbital hold? Schrodinger Wave Equation Y = fn(n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of nandl Orbital – electrons with the same values of n, l, andml If n, l, and mlare fixed, then ms = ½ or - ½ Y = (n, l, ml, ½) or Y = (n, l, ml, -½) An orbital can hold 2 electrons

  27. How many electrons can be placed in the 3d subshell? n=2 n=3 l = 1 l = 2 How many 2p orbitals are there in an atom? If l = 1, then ml = -1, 0, or +1 2p 3 orbitals If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e-

  28. Energy of orbitals in a single electron atom ( ) En = -RH 1 n2 n=1 n=2 n=3 Energy only depends on principal quantum number n

  29. Energy of orbitals in a multi-electron atom n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 Energy depends on n and l

  30. ? ? “Fill up” electrons in lowest energy orbitals (Aufbau principle) Li 3 electrons Be 4 electrons B 5 electrons C 6 electrons B 1s22s22p1 Be 1s22s2 Li 1s22s1 H 1 electron He 2 electrons He 1s2 H 1s1

  31. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 6 electrons N 7 electrons O 8 electrons F 9 electrons Ne 10 electrons Ne 1s22s22p6 C 1s22s22p2 N 1s22s22p3 O 1s22s22p4 F 1s22s22p5

  32. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

  33. What is the electron configuration of Mg? What are the possible quantum numbers for the last (outermost) electron in Cl? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml= -1, 0, or +1 ms = ½ or -½

  34. Outermost subshell being filled with electrons

  35. 2p 2p Paramagnetic Diamagnetic unpaired electrons all electrons paired

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